Trig identities

**imthatgirl** · January 10th 2007, 04:51 PM

can anyone help with this problem?
Prove

tan^2 + cos^2 + sin^2 = 1/cos^2

**ThePerfectHacker** · January 10th 2007, 05:34 PM

Originally Posted by imthatgirl

can anyone help with any of these problems?
Prove
a) (tan^2) /( 1+tan^2) = sin^2

$1+\tan^2 x=\sec ^2 x$
Thus,
$\frac{\tan ^2 x}{\sec^2 x}$
$\frac{(\sin^2 x/\cos^2 x)}{(1/\cos^2 x)}$
Simplify fraction,
$\sin^2 x$

**ThePerfectHacker** · January 10th 2007, 05:50 PM

Originally Posted by imthatgirl

b) tan^2 - cos^2 = 1/(cos^2) - 1 - cos^2

$\tan^2 x - \cos^2 x= \frac{1}{\cos ^2 x}- 1 -\cos ^2 x$

$(\sec^2 x-1)-\cos^2 x=\frac{1}{\cos^2 x}-1-\cos^2 x$

$\sec^2 x - 1 - \cos^2 x= \frac{1}{\cos^2 x}-1 -\cos^2 x$

$\frac{1}{\cos^2 x}-1-\cos^2 x = \frac{1}{\cos^2 x}-1 -\cos^2 x$

**imthatgirl** · January 10th 2007, 06:21 PM

thank you so much for the help i get it now

**earboth** · January 10th 2007, 10:30 PM

Originally Posted by imthatgirl

can anyone help with this problem?
Prove

tan^2 + cos^2 + sin^2 = 1/cos^2

Hello,

you have to use these properties:

$\cos^2(x)+\sin^2(x)=1$
$\tan(x)=\frac{\sin(x)}{\cos(x)}$

$\tan^2(x)+\cos^2(x)+\sin^2(x)=\tan^2(x)+1-\sin^2(x)+\sin^2(x)=$ $<br /> \tan^2(x)+1=\frac{\sin^2(x)}{\cos^2(x)}+1=\frac{\s in^2(x)+\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}$

EB

PS: If you have a new problem please start a new thread. Otherwise you risk that nobody will notice that you asked for some help.

**topsquark** · January 11th 2007, 03:56 AM

I'll reinforce that comment. If you remove your original question then no one can see what you originally posted and thus be confused about what someone's response to it is. If someone else is looking for help on a similar topic, that makes it harder for them to learn it themselves. So please, when you have a new question either post a completely new thread, or just respond to the post you already started.

-Dan

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